[Math] A canonical and categorical construction for geometric realization

at.algebraic-topologyct.category-theorysimplicial-stuff

There is a very intimate connection between categories, simplicial sets, and topological spaces. On one hand, simplicial sets are the presheaf category on the category $\Delta$ and $\Delta$ is a canonically defined "invariant" of the theory of categories. (e.g. the machinery of Mark Weber "spits out" $\Delta$ when you "plug in" the free category monad: http://golem.ph.utexas.edu/category/2008/01/mark_weber_on_nerves_of_catego.html)

However, $\Delta$ is also linked with topological spaces. The key to this link is the functor $\Delta \to Top$ which assigns the category $[n]$ the standard n-simplex $\Delta^n$. It is this functor which produces the adjunction between the geometric realization functor and the singular nerve functor which allow you to transfer the model structure on $Top$ to $Set^{\Delta^{op}}$ so that this adjunction becomes a Quillen equivalence.

My question is the following:

Is there a deep categorical justification for the functor $\Delta \to Top$ being defined exactly how it is? If we didn't know about the standard n-simplices, how could we "cook up" such a functor? I would like a construction of this functor which is truly canonical.

The closest to an answer I've found is Drinfeld's paper http://arxiv.org/abs/math/0304064. However, this doesn't quite "nail it home" to me. First of all, the definition is just made, but not motivated. The definition shouldn't be a "guess that works", but something canonical. Moreover, if you unwind it enough, it is secretly using the fact that finite subsets of the interval with cardinality $n$ correspond to points in (the interior of) the $(n+1)$-simplex. Plus, there's some funny business going on for geometric realization of non-finite simplicial sets. (Don't get me wrong- I think it's a great paper. It just doesn't totally answer my question).

EDIT: A possible lead:

$Set^{\Delta^{op}}$ is the classifying topos for interval objects and the standard geometric realization functor $Set^{\Delta^{op}} \to Top$ is uniquely determined by its sending the generic interval to $[0,1]$. This reduces the question to "why is [0.1] the canonical interval?". Is it perhaps the unique interval object whose induced functor $Set^{\Delta^{op}} \to Top$ is both left-exact and conservative?

EDIT: I've proposed a partial answer to this below, along the lines of the above lead. I would love any feedback that anyone has on this.

Best Answer

As to "why is the unit interval the canonical interval?", there is an interesting universal property of the unit interval given in some observations of Freyd posted at the categories list, characterizing $[0, 1]$ as a terminal coalgebra of a suitable endofunctor on the category of posets with distinct top and bottom elements.

There are various ways of putting it, but for the purposes of this thread, I'll put it this way. Recall that the category of simplicial sets is the classifying topos for the (geometric) theory of intervals, where an interval is a totally ordered set (toset) with distinct top and bottom. (This really comes down to the observation that any interval in this sense is a filtered colimit of finite intervals -- the finitely presentable intervals -- which make up the category $\Delta^{op}$.) Now there is a join $X \vee Y$ on intervals $X$, $Y$ which identifies the top of $X$ with the bottom of $Y$, where the bottom of $X \vee Y$ is identified with the bottom of $X$ and the top of $X \vee Y$ with the top of $Y$. This gives a monoidal product $\vee$ on the category of intervals, hence we have an endofunctor $F(X) = X \vee X$. A coalgebra for the endofunctor $F$ is, by definition, an interval $X$ equipped with an interval map $X \to F(X)$. There is an evident category of coalgebras.

In particular, the unit interval $[0, 1]$ becomes a coalgebra if we identify $[0, 1] \vee [0, 1]$ with $[0, 2]$ and consider the multiplication-by-2 map $[0, 1] \to [0, 2]$ as giving the coalgebra structure.

Theorem: The interval $[0, 1]$ is terminal in the category of coalgebras.

Let's think about this. Given any coalgebra structure $f: X \to X \vee X$, any value $f(x)$ lands either in the "lower" half (the first $X$ in $X \vee X$), the "upper" half (the second $X$ in $X \vee X$), or at the precise spot between them. Thus, you could think of a coalgebra as an automaton where on input $x_0$ there is output of the form $(x_1, h_1)$, where $h_1$ is either upper or lower or between. By iteration, this generates a behavior stream $(x_n, h_n)$. Interpreting upper as 1 and lower as 0, the $h_n$ form a binary expansion to give a number between 0 and 1, and therefore we have an interval map $X \to [0, 1]$ which sends $x_0$ to that number. Of course, should we ever hit $(x_n, between)$, we have a choice to resolve it as either $(bottom_X, upper)$ or $(top_X, lower)$ and continue the stream, but these streams are identified, and this corresponds to the identification of binary expansions

$$.h_1... h_{n-1} 100000... = .h_1... h_{n-1}011111...$$

as real numbers. In this way, we get a unique well-defined interval map $X \to [0, 1]$, so that $[0, 1]$ is the terminal coalgebra.

(Side remark that the coalgebra structure is an isomorphism, as always with terminal coalgebras, and the isomorphism $[0, 1] \vee [0, 1] \to [0, 1]$ is connected with the interpretation of the Thompson group as a group of PL automorphisms $\phi$ of $[0, 1]$ that are monotonic increasing and with discontinuities at dyadic rationals.)